1

CE 366 – BEARING CAPACITY (Problems & Solutions) P1 Question:

An excavation will be made for a ten storey 15x25 m building. Temporary support of

earth pressure and water pressure will be made by deep secant cantilever pile wall. The

gross pressure due to dead and live loads of the structure and weight of the raft is 130 kPa

(assume that it is uniform).

this level can only be fill (placed after attained after a construction is

water proofing will relatively long time fully completed) be provided

10 storey building (15x25m)

2 1m

original GWT 4m position

medium medium

1 sat = 20 kN/m3

moist =18 kN/m3

dense dense GWT is lowered 4m sand sand

medium stiff clay 2m sat = 21 kN/m3

a) What is net foundation pressure at the end of construction but before the void space

between the pile wall and the building has been filled, and there is no water inside the

foundation pit yet (water level at the base level) (GWT position 1). b) What is net foundation pressure long after the completion of the building, i.e. water

level is inside the pile wall and the backfill between the building and the pile wall is

placed (GWT position 2). What is the factor of safety against uplift?

2

f

f

f

o o

Solution: a) qnet = final effective stress - initial effective stress

at foundation level at foundation level

1m moist = 18 kN/m3

5m

sat = 20 kN/m3

‘ ‘ = 18x1 + 4x(20-9.8) = 58.8 kPa

( gross pressure – uplift pressure) = final effective stress at foundation level, ’

gross pressure =130 kPa (given)

uplift pressure = 0 kPa (Since GWT is at foundation level (1), it has no effect on

structure load)

’ =130 –0 = 130 kPa qnet =130 – 58.8

= 71.2 kPa

b) ’

’ o

= 130 – 4x9.8 = 90.8 kPa uplift pressure

= 58.8 kPa (same as above)

qnet = 90.8 – 58.8 = 32.0 kPa

OR

qnet=qgross-satD =130-(18x1+4x20) =32.0 kPa

Factor of safety against uplift is:

(FS)uplift = weight of structure / uplift

= (130x15x25) / (4x9.8x15x25) = 3.3

3

P2 Question:

Calculate the FS against uplift and calculate effective stress at the base level for water

level at (1) and (2) for the canal structure given below. Note that the canal is very long

into the page.

(2)

0.75 3.50 0.75 concrete = 24 kN/m3

5.0 m (2)

3.0

ground level

very long concrete pit

1.0

(1) (1) 3.0 m

waterproof membrane

2.0 1.0

Solution:

water table at (1)

Factor of Safety against uplift = (2x6x0.75 + 5x1)x24 / (3x5)x9.8

weight of pit uplift

= 336 / 147

= 2.28

Base pressure = 336 / 5 = 67.2 kN/m2 due to weight of structure.(per meter of canal)

147 / 5 = 29.4 kN/m2 is supported by groundwater

67.2 – 29.4 = 37.8 kN/m2 is supported by soil (effective stress at the base)

base pressure 29.4 kPa : supported by

due to 67.2 kPa groundwater (uplift)

structure 37.8 kPa : supported by

soil

2.85

4

water table at (2)

FS = 336 / (6.85x5x9.8)

= 1.0 < 1.5 NOT OKEY

⇒ base pressure = 67.2 kPa is supported by ground water

uplift = weight of structure

Soil does not carry any load, structure tends to float

5

P3 Question:

A residential block will be constructed on a clay deposit. The building will rest on a mat

foundation at 2m depth and has 20mx20m dimensions in plan.

The clay deposit is 26m deep and overlies limestone. The groundwater level is at 2m

depth. The bulk unit weights are 18 and 20 kN/m3 above and below water table respectively.

The clay has c’=5 kN/m2, ’=200, cu=48 kN/m2, u=0. The coefficient of volume

compressibility is 1.00x10-4 m2/kN at the ground surface and decreases with depth at a

rate of 0.02x10-4 m2/kN per meter. Use Eu/cu = constant = 1250 and Is = 1.2 a) Calculate ultimate bearing capacity of the foundation in the short term?

b) For the foundation described above what is the (gross) allowable bearing capacity?

NOTE: For u=0 case use Skempton values, use a safety factor of 3.00 against shear

failure of the foundation. Use sublayers. Maximum allowable total settlement of the

building is 15 cm. Solution:

2m 20x20

26m z d=18kN/m3

sat=20kN/m3

c’=5kPa ’=20

cu=48 kPa u

limestone Skempton expression for u is : qf = cuNc + sat D (total stress analysis)

qnf = cuNc

6

Short Term :

D

2 B 20

0.1

N c square

6.4

(Skempton Chart, page 73 Fig.4.6 in Lecture Notes)

q f 48x6.4 18x 2 343.2 kPa

q nf q f D c u N c 307.2 kPa

Settlement Check :

St = Si + Sc

IMMEDIATE SETTLEMENT IN CLAY, Si:

S qB (1 2 )I i E s

where

q q net (net foundation presure) qnf

FS 307.2

102.4 kPa 3

Note that in clay for UNDRAINED CASE 0.5 undrained mod ulus, Eu 60 000 kPa Is 1.2 (given)

S 102.4x20 (1 0.52 )x1.2 0.031m 31mm i 60x103

CONSOLIDATION SETTLEMENT IN CLAY, Sc:

2m 20x20

26m z . mid-point of sublayer 1, z1=6m

. mid-point of sublayer 2, z2=18m

H1=12 m

H2=12 m

limestone

7

Vertical Stress due to qnet should be determined at the mid-point of each sublayer

10m 10m

Soed = mv

=4qIr ; q=qnet=102.4 kPa

mv = [1-0.2(2+z)]x10-4

Layer no z m=n=10/z Ir mv(m2/kN)

1 6 1.67 0.2 81.9 0.84x10-4

2 18 0.55 0.093 38.1 0.6x10-4

Soed= ( 0.84x10-4x81.9x12)+( 0.6x10-4x38.1x12)=0.110m=110mm

St = 31+110 141mm<150mm (allowable) OK. GENERALLY IN CLAY SHEAR FAILURE CONTROLS THE DESIGN,

SETTLEMENT IS NOT CRITICAL. BUT IT SHOULD BE CHECKED ALSO

(qall)net = 102.4 kPa

(qall)gross= 102.4+2x18 = 138 kN/m2

8

P4 Question:

A footing of 4mx4m carries a uniform gross pressure of 300 kN/m2 at a depth of 1.5m in

a sand. The saturated unit weight of the sand is 20 kN/m3 and the unit weight above the

water table is 17 kN/m3. The shear strength parameters are c’=0, ’=320. Determine the factor of safety with respect to shear failure for the following cases;

a) The water table is at ground surface

b) The water table is 1.5m below the surface Solution:

FS (q ult ) net

q net

q nf

q n

q ult D

q gross D

q f D

q n

For square footing:

q f qult 0.4BN 1.2cNc DNq

c' 0 and ' 320 N 26 , Nq 29 (see page 69 Figure 4.3 in Lecture Notes)

jıjıjıjı

D=1.5m

B=4.0m

9

a) q f 0.4B' N ' DNq 0.4x4x(20 10)x 26 (20 10)x1.5x 29 851kPa

qnf q f ' D 851 (20 10)x1.5 836 kPa

qgross 300 kPa

i. qnet 300 20x1.5 270 kPa OR

ii. q net (300 1.5x10) 1.5(20 10) 270 kPa

FS 836

3.1 270

b) q f 0.4B ' N d DN q

0.4x4x(20 10)x26 17x1.5x29 1156 kPa

q nf q f D 1156 17x1.5 1130 kPa

q gross

q net

300 kPa

300 17x1.5 275 kPa

FS 1130 4.1

275

10

P5 FOOTING ON SAND

Question:

The column loads, wall loads and the pertinent soil data for a proposed structure is given

below.

i. Design the square column and wall footings for a permissible settlement of 30 mm,

using Peck & Hanson & Thornburn charts. Make a reasonable assumption to obtain an

average N value below the footing.

depth 1 2 3 4 5 6 7 8 9 10 11 12 N

8 14 11

16

18

11

9

13

18

20 50/11 50/7

280 kN/m 900 kN 900 kN 280 kN/m

3 m 3 m 3 m

Df =1m

Dw

wall

GWT

column

= 18 kN/m3 1.5m

SAND = 21 kN/m3

w = 10 kN/m3

Footing on Cohesionless Soils:

Assumptions:

significant depth: 0.5 B above, 2 B below the footing

weight of excavated soil weight of (footing + column) in the soil

column load / area qnet

footings to be designed for the largest qnet (i.e. column ftg)

11

epth Nfield ‘ o

CN Ncor 1 8 18 2.0 16 2 14 36 1.63 23 3 11 50.5 1.38 15 4 16 61.5 1.25 20 5 18 72.5 1.15 21 6 11 83.5 1.07 12 7 9 94.5 1.01 9 8 13 105.5 0.95 12 9 18 116.5 0.91 16 10 20 127.5 0.87 17 11 50/11 - 12 50/7 -

Depth

1

Ncor

16 2 23 3 15 4 20 5 21 6 12 7 9 8 12 9 16

Solution: NOTE: For Peck-Hanson-Thorburn, N values should be corrected for overburden stress

D

CN (overburden correction) values are calculated by using eq.2.3 (page 31) in Lecture Notes

i )

Square column footings Peck & Hanson & Thornburn charts:Fig 4.8 in Lecture Notes ⇒ assume B=3.0 m ⇒ To obtain the average N value to be used in the calculations

Consider 0.5B=0.5x3=1.5m above

2.0B=2.0x3=6.0m below the foundation level

0.5B=1.5m

2.0B=6.0m

10 17

Nav = (16+23+15+20+21+12+9) / 7 = 17

12

⇒ assume

Depth

B = 2.0 m

Ncor 1 16 2 23 3 15 4 20 5 21 6 12 7 9 8 12 9 16

Cw = 0.5 + 0.5x[2.5/(1+3)] = 0.81

(qn )all=11 N cw (kN/m2) for 25 mm settlement (page 78 in Lecture Notes)

(qn )all=11x17x0.81 = 151 kPa

(qn ) all

(qn ) all x

Sall (mm) 25

qall = 151x(30/25) = 181 kPa

qnet = 900/(3x3) = 100 kPa

181 >>100 overdesign

0.5B=1.0m

2.0B=4.0m

10 17

Nav = (16+23+15+20+21) / 5 = 19

Cw = 0.5 + 0.5x[2.5/(1+2)] = 0.92

(qn )all=11x19x0.92 = 192 kPa

qall = 192x(30/25) = 230 kPa

qnet = 900/(2x2) = 225 kPa

230 225 OK

B = 2.0 m

13

Wall footings ⇒ Use qnet = 225 kPa

B 280

1.25m 225

Check B value

Nav = (16+23+15) / 3 = 18

No GWT correction

(qn )all=11x18 = 198 kPa qall = 198x(30/25) = 238 kPa

238 > 225 OK

14

P6 FOOTING ON CLAY Question:

A public building consists of a high central tower which is supported by four widely

spaced columns. Each column carry a combined dead load and representative sustained

load of 2500 kN inclusive of the substructure (gross load). Trial borings showed that

there is a 7.6m of stiff fissured Ankara clay (cu=85 kPa, Eu = 30 MN/m2 and

mv = 1x10-4 m2/kN) followed by dense sand. Determine the required foundation depth and

allowable bearing pressure for the tower footings.

Assume wet = sat = 18.6 kN/m2 (above and below GWT)

w = 10 kN/m2

Consider immediate and consolidation settlements. Divide the clay layer into 4 equal

sublayers. The foundation depth can be taken as 2m.

⇒ D=2.0m, cu = 85 kPa Solution:

Assume B=2.0m Df/B=1 ⇒ Nc = 7.7 (Skempton)

qnf = (qult)net = cuNc = 85x7.7 = 654.5 kPa for FS=2.5 (qnet)safe = 654.5/2.5 = 261.8 kPa

1.2

2500kN qnet = 2500/(2x2) – 2x18.6 = 587.8 kPa

OR 2m qnet = (2500/(2x2) – 0.8x10)-(1.2x18.6+0.8x8.6)

= 587.5 kPa

(qnet)safe << qnet NOT ACCEPTED Assume B=3.0m

15

Df/B=0.67 ⇒ Nc = 7.4 (Skempton)

qnf = (qult)net = cuNc = 85x7.4 =629 kPa

for FS=2.5 (qnet)safe = 629/2.5 = 251.6 kPa

qnet = 2500/3x3 – 2x18.6 = 241 kPa

(qnet)safe qnet OK

B=3.0m Settlements B=3.0m Eu = 30000 kPa Df=2.0m

Compressible layer thickness H=7.6-2=5.6m

H 1.87

B D 0.67

B

⇒ 0 0.95 1 0.57

S 0.57x0.95x 241x3 0.013m 13mm

i 30000

2500kN

1.2 2m

1.4

1.4

1.4

1.4

P1

P2

P3

P4

Sand is incompressible (also = 2B)

SAND

qnet=241 kPa

16

))(( zLzBBLqP net

(Use 2:1 approximation)

Layer no Thickness,H (m) P

1

2

3

4

1.4

1.4

1.4

1.4

158

83.4

51.3

34.8

Note that: ⇒ P= vertical stress due to qnet at the mid-point of each sublayer

Soed=mv..H

Soed=1x10-4x1.4x(158+83.4+51.3+34.8)=4.585x10-2m=45.85mm

Apply Skempton-Bjerrum factor =0.5

Sc = Soed 45.85x0.5 22.9mm

Stotal = SI + Sc = 13+22.9 = 35.9mm

17

P7 RAFT FOUNDATION ON DEEP CLAY LAYER Question:

A 16-storey apartment block is to be constructed at a site. The soil profile consists of a

deep clay layer which contains a 5m thick sand layer. The ground water table is at 4m

depth. The base of the raft under the building is 8m deep from the ground surface. The

profile and the soil properties are shown in the figure below.

The dimensions of the building and the raft are the same (15mx30m). Total weight of the

building (dead+live+raft) is 90 000 kN.

Find the net foundation pressure and check the factor of safety against bearing capacity

and calculate the total settlement of the building.

No secondary settlements are expected. Take the Skempton-Bjerrum correction factor =

0.75. Consider the compressions of the soil within 20m distance from the foundation

level.

B=15m Original GWT position

=18 kN/m3

sat=20kN/m3

Medium stiff clay mv1=0.025x10-2

cu=40kN/m2 Eu=20MPa

RAFT

4m

4m I 5m II

Stage 2 Stage 1

5m

Medium stiff clay mv2=0.015x10-2

cu=45kN/m2 Eu=20MPa

III 5m

IV 5m

Total weight of the building (dead+live+raft)=Qgross=90000 kN

qgross= 90000/(15x30) = 200 kPa

18

o

Solution: Stage 1 (GWT is lowered to the foundation level)

Uplift = 0

’=4x18+4(20-9.8) = 112.8 kPa

qnet=(200-0)-112.8 = 87.2 kPa (net foundation pressure) Stage 2 (GWT is raised to its original position)

Uplift= 4x9.8= 39.2 kPa

’ o =4x18+4(20-9.8) = 112.8 kPa

qnet=(200-39.2)-112.8 = 48 kPa

qnet= 87.2 kPa is MORE CRITICAL Net bearing capacity of the foundation : qnf = qf - D=cuNc+D-D=cuNc

cu = 40 kPa

Df/B=8/15=0.53 (Nc)square=7.1

(Nc)rect.=(Nc)square (0.84+0.16B/L) = 7.1(0.84+0.16x15/30) = 6.5

qnf = 6.5x40 = 260 kPa Safety factor against shear

FS q nf

q net

260 87.2

3.0 OK

Settlement Analysis:

Total settlement= St= SI + Sc

Consider the compressions of the soil within 20m distance from the foundation level.

Initial settlement : Si

0 1

qB

0.95x0.5x 87.2x15 3.1cm

E u 20000

19

Consolidation Settlement : Sc = mv H

For consolidation settlement; consider 5m thick sublayers.

15

7.5 = 4qIr

qnet = 48 kPa sınce consolıdatıon is a LONG TERM situation

n=B/z m=L/z Ir = 4qIr mv (m2/kN) 7.5/2.5 15/2.5 0.245 47 0.025x10-2

7.5/7.5 15/7.5 0.2 38.4 0.025x10-2

7.5/12.5 15/12.5 0.145 27.8 0.015x10-2

7.5/17.5 15/17.5 0.102 19.6 0.015x10-2

Sc = 0.025x10-2x47x5 + 0.025x10-2x38.4x5 + 0.015x10-2x27.8x5 + 0.015x10-2x19.6x5

Sc = 0.142m=14.2cm ⇒ =0.75 (Skempton-Bjerrum)

Sc = 14.2x0.75= 10.7cm

St = 3.1+10.7 = 13.8 cm